Question

A basketball player, standing near the basket to grab a rebound, jumps 68.4 cm vertically. How much time does the player spend in the top 12.2 cm of his jump? How much time does the player spend in the bottom 12.2 cm of his jump?

Answer 1

given

h = 68.4 cm

= 0.684 m

g = 9.8 m/sec²

a )

the player spend in the top 12.2 cm

= 0.122 m

using v = ( 2 g h )^1/2

= ( 2 x 9.8 x 0.684 )^1/2

v = 3.6614 m/sec

V₁ = ( 3.6614² - ( 2 x 9.8 x ( 0.684 - 0.122 )) )^1/2

= ( 3.6614² - ( 2 x 9.8 x 0.562 ) )^1/2

= ( 13.4 - 11.01 )^1/2

V₁ = 2.39 m/sec

t₁ = 2 x 0.122 / 2.39

= 0.102 sec

t = 2 t₁

= 2 x 0.102

t = 0.2041 sec

or

so the player spend in the top 12.2 cm of his jump is t = 204.184 m sec

b )

0.122 = 3.6614 x t₂ - 0.5 x 9.8 x t₂²

3.6614 x t₂ - 0.5 x 9.8 x t₂² - 0.122 = 0

4.9 t₂² - 3.6614 t₂ + 0.122 = 0

t₂ = 0.0349 sec

so the time of t₂ = 34.9 m sec the player spend in the bottom 12.2 cm of his jump.